# Inverse of funtion.

1. Oct 5, 2012

### trixitium

I'm reading the first chapter of Topology by Munkres. There we can see:

"if $f$ is bijective, there exists a function from B to A called the inverse of $f$.

(...)

As another situation where care is needed, we note that it is not in general true that

$f^{-1}(f(A_0) = A_0$ and $f(f^{-1}(B_0)) = B_0$. The relevant rules, which we leave you to check, are the following: If $f: A \rightarrow B$ and $A_0 \subset A$ and $B_0 \subset B$, then

$A_0 \subset f^{-1}(f(A_0))$ and $f(f^{-1}(B_0) \subset B_0$

The first inclusion is equality if $f$ is injective and the second inclusion is equality if $f$ is surjective."

Are there any sense in talking about inverse considering that $f$ is not injective or surjective???

2. Nov 2, 2012

### Latrace

Ah, this is a common misunderstanding, due to a notation issue. If $f : A \rightarrow B$ is any function and $B_0 \subset B$, then by $f^{-1}(B_0)$ people always mean the set $\{ x \in A | f(x) \in B_0 \}$. Notice that this has meaning even if $f$ is not a bijection. Likewise, if $A_0 \subset A$, we have $f(A) = \{ f(x) | x \in A \}$. I think that with these definitions you should be able to understand the conclusions made in your book.

3. Nov 2, 2012

### HallsofIvy

Staff Emeritus
Oh, dear, oh, dear! Just seeing this question makes me want to hide under the bed!

The very first time I had to give an explanation of a proof to a class in a topology class, it involved $f^{-1}(A)$ for A a set and I did the whole thing assuming f was invertible!

If f is a function from set X to set Y, and A is a subset of X, B a subset of Y, then we define f(A) to be the set of all y in Y such that f(x)= y for some x in A and $f^{-1}(B)$ to be the set of all x in X such that f(x) is in B.

IF f is "one to one and onto", that is, if f is invertibe, then we can show that $f^{-1}(f(A))= A$, but f does not have to be invertible, or even defined on set B for $f^{-1}(B)$ to be defined.

For example, let f:R=>R be defined by f(x)= x2 and let B= [-4, 4]. Then $f^{-1}(B)= [-2, 2]$. f(2)= f(-2)= 4 so both 2 and -2 are in $f^{-1}(B)$ and for any x between -2 and 2, -4< 0< f(x)< 4, so x is also in $f^{-1}(B)$. If x< -2 or x> 2, f(x)> 4 so not in [-4, 4].

Even $f^{-1}([-4, -1])$ is defined. Because there is NO x such that $f(x)= x^2$ is in [-4, -1] so $f^{-1}([-4, -1])$ is the empty set.